Not certain if this is an appropriate for an answer so I'll leave it as a comment for now. Here's another nice-but-surprising way to get $\omega^\omega$: Let f(n) denote the smallest number of 1's needed to write n using any combination of addition and multiplication, e.g., f(7)=6 as shortest way for 7 is 7=(1+1+1)(1+1)+1. For any n, f(n)>=3log_3(n). So subtract off this lower bound and consider d(n)=f(n)-3log_3(n). Then the set of all values of d is a well-ordered set of real numbers, with order type $\omega^\omega$. Afraid I can't give a proper reference as we haven't published this yet! :)
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Harry AltmanJun 13 '11 at 0:18

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@Harry: your results are definitely interesting and also arise from a natural number theory question, so I'd like to vote it up if you post it as an answer:) By the way, there are two more examples became known to me recently: the order type of the set of Pisot numbers, and the $\sigma$-ordering on braid groups.
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Junyan XuJun 15 '11 at 5:38

2 Answers
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Alright, I'll put my comment as an answer and hopefully get this off the no-upvoted-answers queue. :)

Here's another nice-but-surprising way to get $\omega^\omega$: Let $\|n\|$ denote the smallest number of 1's needed to write n using any combination of addition and multiplication, e.g., $\|7\|=6$ as shortest way for 7 is $7=(1+1+1)(1+1)+1$. (This is known as the "integer complexity" of n; it's sequence A005245.)

Now, for any n, we have the lower bound $\|n\|\ge 3log_3 n$. So subtract this off and consider $\delta(n):=\|n\|-3log_3 n$. Then the set of all values of $\delta$ is a well-ordered subset of $\mathbb{R}$, with order type $\omega^\omega$.

Hrbacek, following Ballard, has recently put a certain partial ordering called $\sqsubseteq$ on absolutely everything in order to do nonstandard analysis a la Nelson (i.e. internal set theory):

Let $\mathscr{L}$ be the language of ZFC (and Tarski-Grothendieck if you insist). We say a well-formed formula of $\mathscr{L}$ is an $\in$-formula. We assert that ZFC holds for all well-formed $\in$-formulae.

Now we throw in $\sqsubseteq$ to $\mathscr{L}$ to get a bigger language $\mathscr{HB}.$ First, we assume that ZFC holds for all $\in$-formulae. Let us write
$x \sqsubseteq_\alpha y$ as an abbreviation of
$x \sqsubseteq \alpha \vee x \sqsubseteq y.$
Suppose we have a well-formed formula $P$ of $\mathscr{HB}.$ We write $P^\alpha$ for the replacement of every instance of $\sqsubseteq$ with $\sqsubseteq_\alpha.$
Let us also write $x \sqsubset y$ for
$x \sqsubseteq y \wedge y \not\sqsubseteq x.$
We also write $2^A_{\mathrm{fin}}$ for the set of all finite subsets of $A.$
We also write $(\forall u \sqsubseteq v) P(u,v)$ for $(\forall u)(u \sqsubseteq v \implies P(u,v)),$ and so on for $\exists,$ and for $\in$ as well.

Then Hrbacek's GRIST is the following condition on $\sqsubseteq,$ with four axiom schemata:

R elativization condition on $\sqsubseteq$: $\sqsubseteq$ is a total dense preordering with minimal element
$\emptyset$ and no maximal element; i.e. the conjunction of